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Sonbull [250]
3 years ago
8

Inverse of the function h(x)=\dfrac{3}{2}(x-11)h(x)= 2 3 ​ (x−11)h, left parenthesis, x, right parenthesis, equals, start fracti

on, 3, divided by, 2, end fraction, left parenthesis, x, minus, 11, right parenthesis?
Mathematics
2 answers:
defon3 years ago
6 0

Answer:

h^{-1}(x)=\frac{2}{3}x+11

Step-by-step explanation:

The given function is

h(x)=\dfrac{3}{2}(x-11)

We need to find the inverse of the given function.

Step 1: Substitute h(x)=y.

y=\dfrac{3}{2}(x-11)

Step 2: Interchange x and y.

x=\dfrac{3}{2}(y-11)

Step 3: Isolate y.

\frac{2}{3}x=y-11

\frac{2}{3}x+11=y

Step 4: Interchange sides.

y=\frac{2}{3}x+11

Step 5: Substitute y=h^{-1}(x).

h^{-1}(x)=\frac{2}{3}x+11

Therefore, the inverse of the function is h^{-1}(x)=\frac{2}{3}x+11.

blagie [28]3 years ago
5 0

Answer:

The inverse function of h(x) = 3/2*(X-11)  is:  g(x)= (2/3*X) +11

Step-by-step explanation:

h(x) = Y= 3/2*(X-11)

To find the inverse function, the first step is to exchange the position of the variables with each other. So;

X= 3/2*(Y-11)

Now we need to isolate the variable Y from this equation. The final result will be te inverse function.

X=3/2*(Y-11)

2*X=3*(Y-11)

2/3*X=Y-11

2/3*X +11 = Y

Y= (2/3*X) + 11  (Inverse function)

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Answer:

Final answer is A. P(A^c)=\frac{4}{7}

Step-by-step explanation:

From table we see that there are 7 places in the list.

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3 years ago
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Mazyrski [523]

Answer:

23\sqrt{3}\ un^2

Step-by-step explanation:

Connect points I and K, K and M, M and I.

1. Find the area of triangles IJK, KLM and MNI:

A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\

2. Note that

A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2

3. The area of hexagon IJKLMN is the sum of the area of all triangles:

A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2

Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.

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Answer:

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<A = 45°

< B = x + 65° = 45 °+ 65 °= 110°

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Hope it helps :)

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